\(n+1=\dfrac{n+1}{n+2}\)

=> \(\left(n+1\right)\left(n+2\right)=n+1\)

\(n^2+3n+2=n+1\)

\(n^2+2n+1=0\)

\(\left(n+1\right)^2=0\)

\(n+1=0\)

n=-1

\(\left(x^2+x\right)\left(x^2+x+1\right)=0\)

\(x\left(x+1\right)\left(x^2+x+1\right)=0\)

=> x=0 hoặc x+1=0   (do \(x^2+x+1\ne0\))

                      x=-1

Vậy...

từ bạn ơi sai đấy

\(A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{127}\)

   \(=\left(\dfrac{1}{127}+1\right).127\div2\)

   \(=\dfrac{128}{127}.127\div2\)

   \(=128\div2\)

   = 64

Vậy A>4

a) Xét \(\Delta ABD\) vuông tại A  và \(\Delta HBD\) vuông tại H có :

    BD chung 

     \(\widehat{ABD}=\widehat{HBD}\)

=> \(\Delta ABD\) = \(\Delta HBD\) (ch-gn)

=> AD=HD

b) Xét \(\Delta AKD\) vuông tại A và \(\Delta HCD\)  vuông tại H có :

AD=HD

\(\widehat{ADK}=\widehat{HDK}\)

=> \(\Delta AKD\) = \(\Delta HCD\) (cgvkgn)

=>KD=CD

Xét \(\Delta ADK\) vuông tại A có :

DK là cạnh huyền 

=> DK>AD

=> DC> AD

c) Có : AB=HB ( \(\Delta ABD\) = \(\Delta HBD\))

            AK=HC ( \(\Delta AKD\) = \(\Delta HCD\) )

=> AB+AK=HB+HC

=> BK=BC

=> \(\Delta BKC\) cân tại B

c)Các góc \(35^o\) là : \(\widehat{x'Ay'},\widehat{xAy}\)

d)Các góc \(145^o\) là : \(\widehat{xAy'}\) ,\(\widehat{yAx'}\)

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=5950\)

\(x+x+1+x+2+...+x+99=5950\)

\(100x+\left(1+2+3+...+99\right)=5950\)

\(100x+\left(99+1\right).99:2=5950\)

\(100x+4950=5950\)

\(100x=5950-4950\)

100x=1000

x=10

a)Xét  \(\Delta ABC\) vuông tại A có:

 \(BC^2=AC^2+AB^2\)

\(BC^2=64+36\)

\(BC^2=100\)

BC=10cm

Xét \(\Delta ABC\) có: AD là phân giác của\(\widehat{BAC}\)

=> \(\dfrac{BD}{AB}=\dfrac{DC}{AC}=\dfrac{BD+DC}{AB+AC}=\dfrac{BC}{AB+AC}=\dfrac{10}{6+8}=\dfrac{5}{7}\)

=> \(\dfrac{BD}{AB}=\dfrac{5}{7}\Leftrightarrow\dfrac{BD}{6}=\dfrac{5}{7}\Rightarrow BD=\dfrac{5}{7}.6\approx4,3\) cm

b)

Xét \(\Delta HBA\) và \(\Delta ABC\) có:

 \(\widehat{B}\) chung

 \(\widehat{BHA}=\widehat{BAC}\left(=90^o\right)\)

=> \(\Delta HBA\sim\)\(\Delta ABC\) (g-g)

=> \(\dfrac{AH}{AC}=\dfrac{AB}{BC}\Leftrightarrow\dfrac{AH}{8}=\dfrac{6}{10}\Rightarrow AH=\dfrac{3}{5}.8=4,8cm\)

    \(\dfrac{HB}{AB}=\dfrac{AB}{BC}\Leftrightarrow\dfrac{HB}{6}=\dfrac{6}{10}\Rightarrow HB=\dfrac{3}{5}.6=3,6cm\)

c) Có : \(\dfrac{HB}{AB}=\dfrac{AB}{BC}\) ( \(\Delta HBA\sim\)\(\Delta ABC\) )

=> \(AB^2=HB.BC\)